Chapter 21, Problem 74AE

(a)

Interpretation Introduction

Interpretation:

The complex ion, the counter ions, the electronic configuration of the transition metal and geometry of complex ion for each of the given coordination compounds is to be stated.

Concept introduction:

The electrons in the d orbital of a transition metal split into high and low energy orbitals when ligands are attached to it.  The energy difference between these two levels depends upon the properties of both metal and the ligands.  If the ligand is strong, then splitting will be high and the complex will be low spin.  If the ligand is weak, then splitting will be less and the complex will be high spin.

To determine: The complex ion, the counter ions, the electronic configuration of the transition metal and geometry of complex ion for the given coordination compound.

Answer to Problem 74AE

The complex ion present is ${\left[\text{Co}{\left({\text{H}}_{\text{2}}\text{O}\right)}_6\right]}^{2+}$.  The counter ions are ${\text{2Cl}}^{-}$.  Electronic configuration $\left[\text{Ar}\right]{\text{3d}}^{\text{7}}$.  The geometry of the complex ion is octahedral.

Explanation of Solution

The given compound is ${\text{CoCl}}_{\text{2}}{\text{.6H}}_{\text{2}}\text{O}$. It is used in novelty devices that predict rain.

The coordination compound is $\left[\text{Co}{\left({\text{H}}_{\text{2}}\text{O}\right)}_6\right]{\text{Cl}}_{\text{2}}$.

The bracket indicates the composition of complex ion; hence, the complex ion present is ${\left[\text{Co}{\left({\text{H}}_{\text{2}}\text{O}\right)}_6\right]}^{2+}$.

The counter ions are ${\text{2Cl}}^{-}$.

The oxidation state of cobalt is assumed to be $x$ and it is calculated by the formula,

$\left[\begin{array}{l}\left(\text{x×Number of Co atoms}\right)\text{+}\\ \left({\text{Number of H}}_{\text{2}}{\text{O ligands×Charge on one H}}_{\text{2}}\text{O ligand}\right)\end{array}\right]\text{=Charge on compound}$

The number of water ligands is $6$ and charge on them is zero since water is neutral.  The number of cobalt atoms is $1$.  The charge on compound is $+2$ to balance the $-2$ charge due to two chlorine ligands.

Substitute the values of number of cobalt atoms, water ligands and charge on them in the above equation,

$\begin{array}{c}\left[\left(\text{x×1}\right)\text{+}\left(\text{6×0}\right)\right]\text{=+2}\\ \text{x=+2}\end{array}$

Therefore, oxidation state of cobalt is $+2$.

Electronic configuration of $\text{Co}$ is $\left[\text{Ar}\right]{\text{3d}}^{\text{7}}{\text{4s}}^{\text{2}}$.  On removal of two electrons, it becomes ${\text{Co}}^{\text{+2}}$ with electronic configuration $\left[\text{Ar}\right]{\text{3d}}^{\text{7}}$.  There are three unpaired electrons present in $\text{d}$ orbital.

The $\text{4s,4p,4d}$ orbital intermix to form ${\text{sp}}^{\text{3}}{\text{d}}^{\text{2}}$ hybridization.

Therefore, the geometry of the complex ion is octahedral.

(b)

Interpretation Introduction

Interpretation:

The complex ion, the counter ions, the electronic configuration of the transition metal and geometry of complex ion for each of the given coordination compounds is to be stated.

Concept introduction:

The electrons in the d orbital of a transition metal split into high and low energy orbitals when ligands are attached to it.  The energy difference between these two levels depends upon the properties of both metal and the ligands.  If the ligand is strong, then splitting will be high and the complex will be low spin.  If the ligand is weak, then splitting will be less and the complex will be high spin.

To determine: The complex ion, the counter ions, the electronic configuration of the transition metal and geometry of complex ion for the given coordination compound.

Answer to Problem 74AE

The complex ion present is ${\left[\text{Ag}{\left({\text{S}}_{\text{2}}{\text{O}}_{\text{3}}\right)}_{\text{2}}\right]}^{3-}$.  The counter ions are $3{\text{Na}}^{+}$.  Electronic configuration $\left[\text{Kr}\right]{\text{4d}}^{\text{10}}$.  The geometry of the complex ion is tetrahedral.

Explanation of Solution

The dissociation reaction of ${\text{Na}}_{\text{3}}\left[\text{Ag}{\left({\text{S}}_{\text{2}}{\text{O}}_{\text{3}}\right)}_{\text{2}}\right]$ is,

${\text{Na}}_{\text{3}}\left[\text{Ag}{\left({\text{S}}_{\text{2}}{\text{O}}_{\text{3}}\right)}_{\text{2}}\right]\to 3{\text{Na}}^{+}+{\left[\text{Ag}{\left({\text{S}}_{\text{2}}{\text{O}}_{\text{3}}\right)}_{\text{2}}\right]}^{3-}$

The bracket indicates the composition of complex ion hence the complex ion present is ${\left[\text{Ag}{\left({\text{S}}_{\text{2}}{\text{O}}_{\text{3}}\right)}_{\text{2}}\right]}^{3-}$.

The counter ions are $3{\text{Na}}^{+}$.

The oxidation state of silver is assumed to be $\text{x}$ and it is calculated by the formula,

$\left[\begin{array}{l}\left(\text{x×Number of Ag atoms}\right)\text{+}\\ \left({\text{Number of S}}_{\text{2}}{\text{O}}_{\text{3}}{\text{ ligands×Charge on one S}}_{\text{2}}{\text{O}}_{\text{3}}\text{ ligand}\right)\end{array}\right]\text{=}\text{Charge on compound}$

The number of ${\text{S}}_2{\text{O}}_3$ ligands is $2$ and charge on each is $-2$.

The number of silver atom is $1$.  The charge on compound is $-3$ to balance three ${\text{Na}}^{\text{+}}$ ligands.

Substitute the values of number of silver atoms, ${\text{S}}_2{\text{O}}_3$ ligands and charge on them in the above equation.

$\begin{array}{c}\left[\left(\text{x×1}\right)\text{+}\left(\text{2×-2}\right)\right]\text{=-3}\\ \text{x=+1}\end{array}$

Therefore, oxidation state of silver is $+1$.

Electronic configuration of $\text{Ag}$ is $\left[\text{Kr}\right]{\text{4d}}^{\text{10}}{\text{5s}}^{\text{1}}$.  On removal of one electron, it becomes ${\text{Ag}}^{+}$ with electronic configuration $\left[\text{Kr}\right]{\text{4d}}^{\text{10}}$.  The $\text{d}$ orbital is filled. The $\text{5s,5p}$ orbitals intermix to form ${\text{sp}}^{\text{3}}$ hybrid orbitals.

Therefore, the geometry of the complex ion is tetrahedral.

(c)

Interpretation Introduction

Interpretation:

The complex ion, the counter ions, the electronic configuration of the transition metal and geometry of complex ion for each of the given coordination compounds is to be stated.

Concept introduction:

The electrons in the d orbital of a transition metal split into high and low energy orbitals when ligands are attached to it.  The energy difference between these two levels depends upon the properties of both metal and the ligands.  If the ligand is strong, then splitting will be high and the complex will be low spin.  If the ligand is weak, then splitting will be less and the complex will be high spin.

To determine: The complex ion, the counter ions, the electronic configuration of the transition metal and geometry of complex ion for the given coordination compound.

Answer to Problem 74AE

The complex ion present is ${\left[\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}\right]}^{+}$.  The counter ions are ${\text{Cl}}^{-}$.  Electronic configuration $\left[\text{Ar}\right]{\text{3d}}^{\text{10}}$.  The geometry of the complex ion is tetrahedral.

Explanation of Solution

The dissociation reaction of $\left[\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}\right]\text{Cl}$ is,

$\left[\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}\right]\text{Cl}\to {\left[\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}\right]}^{+}+{\text{Cl}}^{-}$

The bracket indicates the composition of complex ion hence the complex ion present is ${\left[\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}\right]}^{+}$.

The counter ion is ${\text{Cl}}^{-}$.

The oxidation state of copper is assumed to be $x$ and it is calculated by the formula,

$\left[\begin{array}{l}\left(\text{Number of Cu atoms×x}\right)\text{+}\\ \left({\text{Number of NH}}_{\text{3}}{\text{ ligands×Charge on one NH}}_{\text{3}}\text{ ligand}\right)\end{array}\right]\text{=Charge on compound}$

The number of ${\text{NH}}_{\text{3}}$ ligands is $4$ and charge on the is zero because they are neutral.

The number of $\text{Cu}$ atom is $1$.  The charge on compound is $+1$ to balance negative charge of chloride ligand.

Substitute the values of number of $\text{Cu}$ atom, ${\text{NH}}_{\text{3}}$ ligands and charge on them in the above equation.

$\begin{array}{c}\left[\left(\text{1×x}\right)\text{+}\left(\text{4×0}\right)\right]\text{=-1}\\ \text{x=+1}\end{array}$

Therefore, oxidation state of $\text{Cu}$ is $+1$.

Electronic configuration of $\text{Cu}$ is $\left[\text{Ar}\right]{\text{3d}}^{\text{10}}{\text{4s}}^{\text{1}}$.  On removal of one electron, it becomes ${\text{Cu}}^{\text{+}}$ with electronic configuration $\left[\text{Ar}\right]{\text{3d}}^{\text{10}}$.  The $d$ orbital is filled.  The $\text{4s,4p}$ orbital intermix to form ${\text{sp}}^{\text{3}}$ hybrid orbitals.

Therefore, the geometry of the complex is tetrahedral.